Why Composite Domains Need Two Checks
A composite formula is built by feeding the outputs of one or more functions into another function. Its domain depends on both stages: each inner expression must first be defined, and the values it produces must be acceptable inputs to the outer function.
For example, even if the expressions substituted into a square root are defined for every real input, the square root still requires a nonnegative input. Conversely, an outer function’s input condition cannot rescue an inner expression that is itself undefined.
The reliable approach is therefore to check the composition in order: establish where the inner expressions exist, then impose the outer function’s domain conditions on their outputs.
Composition and Its Domain
Suppose \(g_1,\ldots,g_k\) are inner scalar functions of an input \(\mathbf{x}\), and \(F\) is an outer scalar function of \(k\) inputs. The composite formula is
Here, \(\mathbf{x}\) may represent one real variable or a tuple of several real variables. Each inner function has its own domain, and \(F\) has a domain consisting of allowable \(k\)-tuples.
This definition gives two necessary checks. First, \(\mathbf{x}\) must lie in the domain of every inner function. Second, after evaluating those functions, their outputs must satisfy all restrictions imposed by \(F\). These conditions are combined by intersection: a point must satisfy all of them at once.
For standard formulas, the outer restrictions often come from familiar rules: a square root requires a nonnegative radicand, a logarithm requires a positive argument, and a denominator must be nonzero. In a composition, apply each rule to the expression that actually serves as the outer function’s input.
A Direct Derivation of the Domain Rule
The domain rule follows from the meaning of evaluation. To evaluate the composite at an input \(\mathbf{x}\), one must first compute every inner value \(g_i(\mathbf{x})\). If even one of those values is undefined, there is no tuple to pass to \(F\). If all inner values exist, evaluation can continue only if that tuple is in the domain of \(F\).
Require \(\mathbf{x}\) to be in the domain of each \(g_i\), so the tuple of outputs can be formed.
Require \((g_1(\mathbf{x}),\ldots,g_k(\mathbf{x}))\) to belong to the domain of \(F\), so \(F\) can be evaluated at that tuple.
Failing either check makes the full composite undefined; passing both makes each stage of the evaluation valid.
This argument proves both directions of the rule. If the composite is defined, then its inner values must exist and must be allowable inputs to \(F\). If both conditions hold, the inner values can be computed and then supplied to \(F\), so the composite is defined.
Worked Example: A Square Root of Two Inner Expressions
Worked Example: A Square-Root Composition
Find the domain of
View this as the outer function \(F(u,v)=\sqrt{u-v}\), with inner expressions \(g_1(x,y)=x+y\) and \(g_2(x,y)=x-y\).
Step 1: Check the inner expressions. Both \(x+y\) and \(x-y\) are polynomials, so each is defined for every real pair \((x,y)\). There are no inner restrictions.
Step 2: Check the outer function. The square root requires its input to be nonnegative. The input \(u-v\), after substitution, is
Therefore the outer condition is \(2y\ge 0\), which is equivalent to \(y\ge 0\).
Step 3: Combine the conditions. Since the inner expressions impose no restrictions, the composite is defined exactly when \(y\ge 0\).
Check: At \((x,y)=(3,1)\), the inner values are \(g_1=4\) and \(g_2=2\), so the outer input is \(4-2=2\), which is nonnegative. At \((3,-1)\), the inner values are \(2\) and \(4\), so the outer input is \(2-4=-2\); the square root is not real. This agrees with the condition \(y\ge 0\).
Worked Example: Inner Square Roots and an Outer Denominator
Worked Example: A Quotient of Two Inner Roots
Find the domain of
The inner expressions are \(g_1(x)=\sqrt{x+2}\) and \(g_2(x)=\sqrt{2-x}\). The outer function is \(F(u,v)=1/(u-v)\), defined when \(u-v\ne 0\).
Step 1: Check where both inner expressions exist. For the first square root, \(x+2\ge 0\), so \(x\ge -2\). For the second, \(2-x\ge 0\), so \(x\le 2\). Together, these conditions give \(-2\le x\le 2\).
Step 2: Check the outer denominator. The denominator must not be zero. Within the interval just found, both square roots are nonnegative, so equality of their values is equivalent to equality of their radicands:
Thus \(x=0\) must be excluded. Combining this with the inner restrictions, the domain is \([-2,0)\cup(0,2]\).
Check: At \(x=-2\), the expression is \(1/(0-2)=-1/2\), so the left endpoint is allowed. At \(x=2\), it is \(1/(2-0)=1/2\), so the right endpoint is allowed. At \(x=0\), the denominator is \( \sqrt{2}-\sqrt{2}=0\), so that input is excluded.
Worked Example: A Logarithm and a Square Root Outside
Worked Example: Two Different Outer Restrictions
Find the domain of
This is an outer function \(F(u,v)=\ln u+\sqrt v\) applied to the inner expressions \(g_1(x,y)=x-y\) and \(g_2(x,y)=1-x^2-y^2\). Both inner expressions are polynomials, so they are defined for every real pair.
Step 1: Determine the outer domain conditions. The logarithm requires \(u>0\), while the square root requires \(v\ge 0\). Substituting the inner expressions gives
The second inequality is equivalent to \(x^2+y^2\le 1\). Hence the domain consists of the points satisfying both \(x-y>0\) and \(x^2+y^2\le 1\).
Check an included point: At \((x,y)=(1/2,0)\), the logarithm input is \(1/2>0\), and the square-root input is \(1-(1/2)^2-0^2=3/4\ge 0\). Both outer operations are defined.
Check an excluded point: At \((0,0)\), the square-root input is \(1\), but the logarithm input is \(0\), and \(\ln 0\) is undefined. Satisfying one outer condition does not compensate for failing the other.
Common Pitfall: Making Every Restriction Strict
A frequent error is to treat every square-root condition as a strict inequality. For a real square root, zero is allowed: the correct condition is that the radicand be greater than or equal to zero.
In the quotient example, an incorrect first step would be to require \(x+2>0\) and \(2-x>0\), producing \(-2<x<2\). That wrongly removes \(x=-2\) and \(x=2\). The correction is to use \(x+2\ge 0\) and \(2-x\ge 0\), giving \(-2\le x\le 2\), and then apply the separate outer-denominator condition. Direct substitution confirms that both endpoints give nonzero denominators.
A Practical Order for Domain Checks
For a composite formula, writing the stages separately helps prevent missed conditions. Identify the inner expressions and the outer rule, then translate every rule into a condition on the original variables. Simplify only after all conditions have been recorded.
- For each inner expression, check whether it has a denominator, an even root, a logarithm, or another restriction.
- Write down the outer function’s domain conditions in terms of its own input variables.
- Substitute the inner expressions into those conditions.
- Combine all conditions, preserving equality where it is allowed.
- Test boundary points or simple sample points when they help confirm the result.
This order matters when an inner expression is itself restricted. An outer square root condition, for instance, may be satisfied algebraically at a point where an inner logarithm is undefined. The full composition still has no value there.
Check Your Understanding
For each question, identify the inner and outer restrictions before simplifying the final condition.
- Find the domain of \(\sqrt{x^2-y}\), and state which part of the condition comes from the outer function.
- For \(\ln(\sqrt{x+1})\), what condition must \(x\) satisfy? Explain why the logarithm requires more than a nonnegative square-root input.
- In \(1/(\sqrt{x+3}-1)\), find the inner restrictions and the value excluded by the outer denominator.
- Give one point where \(\ln(x-y)+\sqrt{1-x^2-y^2}\) is defined and one point where it is not. Check both outer conditions at each point.
- In your own words, why is it not enough to check only the domain of the outer function?